getting some trouble
with dis
square root of x^2+6x=x+square root of 2x
$\displaystyle \sqrt{x^2+6x}=x+\sqrt{2x}$
$\displaystyle \sqrt{x^2+6x}=\sqrt{\left(x+\sqrt{2x}\right)^2}$
$\displaystyle \sqrt{x^2+6x}=\sqrt{x^2+2x\sqrt{2x}+2x}$
$\displaystyle 6x=2x\sqrt{2x}+2x$
$\displaystyle 6x-2x=4x=2x\sqrt{2x}$
Do not do the following....
$\displaystyle \frac{4x}{2x}=2=\sqrt{2x}$
$\displaystyle 2=\sqrt{4}=\sqrt{2x}$
as you lose a solution.
Instead, factor...
$\displaystyle 4x-2x\sqrt{2x}=0$
$\displaystyle x\left(4-2\sqrt{2x}\right)=0$
Please use brackets to make you equation unambiguous, the above could plausibly mean either:
square_root(x^2+6x)=x+square_root(2x)
or
square_root(x^2)+6x=x+square_root(2x)
I know that the first of these is the most likely, but someone will always fail to make that assumption and interpret it differently.
CB
It finds the x=2 solution, but not the x=0 solution.
Quadratic equations may be solved by factoring, but factoring when possible, finds all solutions and is not confined to quadratics.
$\displaystyle 2x=x\sqrt{2x}$
$\displaystyle x\left(2-\sqrt{2x}\right)=0$
zero multiplied by another value is zero, so the factored version shows that x=0 is also a solution.
If we divide both sides by x, we lose that particular solution, though it finds the x=2 answer.
Also remember that we really should not divide by x because we cannot divide by zero anyway,
so when we do that, we are finding only non-zero solutions.
Do check CaptainBlack's post also.